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kenseto
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 Posted: Mon Jul 14, 2008 1:33 pm Post subject: Re: The SR definition led physicists to a 100 years of wild |
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On Jul 11, 1:04 pm, PD <TheDraperFam...@gmail.com> wrote:
| Quote: |
On Jul 11, 9:54 am, kenseto <kens...@erinet.com> wrote:
On Jul 10, 11:36 am, PD <TheDraperFam...@gmail.com> wrote:
On Jul 10, 10:31 am, kenseto <kens...@erinet.com> wrote:
On Jul 10, 11:22 am, PD <TheDraperFam...@gmail.com> wrote:
On Jul 10, 10:07 am, kenseto <kens...@erinet.com> wrote:
On Jul 10, 6:43 am, PD <TheDraperFam...@gmail.com> wrote:
On Jul 9, 4:15 pm, kenseto <kens...@erinet.com> wrote:
On Jul 7, 1:23 pm, PD <TheDraperFam...@gmail.com> wrote:
On Jul 7, 10:26 am, kenseto <kens...@erinet.com> wrote:
On Jul 5, 3:03 pm, PD <TheDraperFam...@gmail.com> wrote:
On Jul 5, 7:45 am, kenseto <kens...@erinet.com> wrote:
On Jul 4, 9:47 am, PD <TheDraperFam...@gmail.com> wrote:
No....he said the train observer M' is rushing toward the light front
from the front and thus he sees the this light front first. He also
said that the train observer is recdeing away from the light front
from the rear and thus he sees this light front at a later time. These
assertions is the baisis of RoS.
No it isn't. It is showing that what the train observer SAW is also
consistent with the laws of physics as known by the track observer.
You do not comprehend what you are reading.
Hey idiot....not according to RoS. RoS says that the train observer is
rushing toward the light front from the front (c+v) and receding from
the light front from the rear (c-v). That means that RoS says that the
speed of light in the train is anisotropic.
That is NOT the statement of the relativity of simultaneity. You don't
comprehend what you are reading.
It is a statement explains how RoS is derived. It is a statement says
that the speed of light in the train is anisotropic..
No, it does NOT say the speed of light is anisotropic in the train.
Yes it does. The train observer is rushing toward the light front from
the front (c+v) and he is receding away from the light from the rear
(c-v). That is anisotropic.
The isotropy of light speed does not have anything to do with the
*closing* speeds that have the values (c+v) and (c-v). There is no law
of physics nor any SR that says the *closing speed* is isotropic >or invariant.
The simultaneity of events is also independent of closing speeds. The
gedanken specified the following conditions:
1. The speed of light in the track and the train frame is isotropic.
2. Both the track observer and the train observer are at equal
distance from the strikes.
3. It was also specified that the track observer sees the strikes to
be simultaneous to begin with.
4. From these conditions the train observer must also see the strikes
to be simultaneous.
You do not know the difference between closing speed and measured
light speed.
It would help if you knew the meaning of the terms you are using.
The *closing* speed of c+v that the track observer sees between the
light from the front and the train observer is NOT the approach speed
of the light measured by the train observer.
So why did you use closing speed to determine whether the train
observer sees the strikes to be simultaneous or not???
I don't. The train observer DOES determine that for himself, using
measured approach speed of light (which is c). That is not the point
of looking at the problem from the track frame. Let me repeat: it is
NOT the purpose of looking at the problem from the track frame to
determine whether the strikes are simultaneous in the train frame.
Einstein did....so are you saying that he wa swrong?
No, he did not. You do not comprehend what you are reading.
The
fact that the strikes are not simultaneous in the train frame is
ALREADY determined from the observations by the train observer.
This is bull shit. The train observer cannot make such determination.
Of course he can. He can SEE the flashes of light, you idiot.
Why? Because he does not know if the strikes were simultaneous or not
simultaneous to begin with.
No, he doesn't have to REASON it out. He OBSERVES things.
The
purpose of looking at the problem from the train frame is to show that
the train observer's account is ALSO consistent with the laws of
physics as known by the track observer. Note that simultaneity is not
a law of physics -- one does not need to have the same answer about
simultaneity for the two frames. The principle of relativity is a
statement about the invariance of the *laws of physics* between
inertial frames, not a statement about the invariance of simultaneity
-- simultaneity is not a law of physics.
You mean RoS is not a law of physics???
You do not even comprehend what I wrote. Please read the above and try
again. Jeez, your command of the English language is pathetic..
I said: "Simultaneity is not a law of physics". You said "You mean ROS
is not a law of physics???"
Do you see what kind of bonehead mistakes you are making?
I agree!!!! It is a buch of
bull shit cook up by Einstein to show that time running at different
rates in different frames. \
Don't you think
that the train observer should do that for himself?
The train observer in
fact measures the approach speed of the light to be c, not c+v.
Right....also he said that the speed of light in his frame is
isotropic and that he is at equal distance from the strikes. Therefore
he determines that the strikes were simltaneous to begin with.
No, those two conditions are not sufficient for simultaneity. The
ADDITIONAL requirement is that the observer actually observes the
light arriving at the observer at the same time.
Right....Einstein already specified that the track observer sees the
strikes to be simultaneous to begin with. Since the train observer
meets the same conditions as the track observer then the train
observer must also sees the strikes to be simultaneous.
But he DOESN'T because he has his OWN set of observations. He is not
reliant on the track observer's observations, or the track observer's
arguments.
This condition is not
met by the train observer. To recap, there are THREE conditions that
must be met for two events to be simultaneous:
1. The speed of the signal from the events to the observer must be
isotropic.
2. The observer is midway between the two events.
3. The observer actually observes the signal arriving at his location
at the same time.
The first two are met in this case, but not the third.
Both observers meet the third condition.
No sir. READ the gedanken experiment and try to comprehend what you
are reading. It says in black and white that the train observer does
NOT meet the third condition. He in fact observes the OPPOSITE.
The train observer does not observe whether the strikes are
simultaneous or not....
Yes, he does. READ the gedanken.
he does not have enough data to make such
observation.
Yes, he does. You've already (just yesterday!) asked me what
observations he made, and I told you, and you've already (just
overnight) forgotten what I told you. Your mind is gone, Seto. You
cannot remember a discussion from one day to the next. I'm about to
give up on you, as your senility is too deep a handicap to overcome.
What he observes are as follows:
1. that the strikes happened at equal distance from him.
2. That the speed of light in his frame is isotropic.
And he observes one more thing which you keep forgetting, day after
day after day, even though it is printed in black and white in the
book you refuse to look up.
3. That the light from the two flashes arrives at his eyes at
different times.
But item #3 is based on the bogus assertion that closing speeds as
perceived by the track observer will affect the isotropy of the speed
of light in the train.
No, it's based on what he SEES.
Hey idiot he does not see that the strikes were not simultaneous.
Why yes, yes, he does, and the statement of the gedanken experiment
says that explicitly. You do not comprehend what you read, Ken.
You could try reading it again:http://books.google.com/books?id=3H46AAAAMAAJ,
try pages 31-33. Note what he says EXACTLY.
|
Fucking idiot....why don't you copy and paste the exact statement that
led Einstein to conclude that the light from the strikes arrives at
the train at different times??
Ken Seto
| Quote: |
Ken, I'm not going to repeat this point again. You have reiterated the
same error every day for the last two weeks, been corrected on it
every day, and you repeat the very same erroneous statement the next
day. You do not remember from one day to the next what is said. I am
no longer interested in talking with someone who is so senile that you
cannot remember what you were told yesterday. I also have no
confidence that if you bother to read the link above today, that you
will remember any of it tomorrow.
Einstein said said that he (the train observer) rush toward the light
from the front and reced away the light from the rear and that's why
he must see the strikes arrive at him at different times. The problem
with this assertion is that it violates the measured isotropy of the
speed of light in the train.
This again is a repetition of the same error made every day without
change for the last two weeks. It is impossible for you to learn
something on Wednesday that you will remember on Thursday. I know it
is painful for you, Ken, but you have a reality that has set in that
you need to deal with.
It is not based on any assumption,
argument, or chain of logic. It is not based on a model, it is not
based on a postulate or principle of relativity. It is the result of
DIRECT OBSERVATION. This is what he SEES
...
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paparios@gmail.com
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| Posted: Mon Jul 14, 2008 7:09 pm Post subject: Re: The SR definition led physicists to a 100 years of wild |
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On 14 jul, 09:27, kenseto <kens...@erinet.com> wrote:
| Quote: |
On Jul 10, 11:36 am, PD <TheDraperFam...@gmail.com> wrote:
No, it's based on what he SEES. It is not based on any assumption,
argument, or chain of logic. It is not based on a model, it is not
based on a postulate or principle of relativity. It is the result of
DIRECT OBSERVATION. This is what he SEES with his own EYES.
Hey idiot there is no such observation. Einstein used closing
velocities to conclude that the light from the strikes arrive at the
train observer at different times. Such conclusion is not OBSERVATION
and such conclusion violates the isotropy of the speed of light in the
train. You are so stupid.
Ken Seto
|
OK, I will put here the complete derivation, proving the relativity of
simultaneity. I'll use some numbers, so you can check the results. I'm
using the same notation as Einstein in http://www.bartleby.com/173/9.html
v= 0.6c = 180000 km/sec
c= 300000 km/sec
sqrt(1-v^2/c^2) = 0.8
Let K be the frame of the track observer M. Also let K’ be the frame
of the train observer M’. The Lorentz transformation equations between
these two frames are:
x'=(x-vt) / (sqrt(1- v^2/c^2)) ; t’ = (t-vx/c^2) / (sqrt(1- v^2/c^2))
When, on frame K, (x,t) = (0,0) it follows that, on frame K’, (x’,t’)
= (0,0) , meaning observers M and M’ coincide. At that instant of time
(t = t' =0), there are two strikes at locations A, at x_A = -100000km,
and location B, at x_B= + 100000km
The back strike light signal moves along the track frame K as x = ct.
Substituting x in Lorentz equations, we obtain:
x'=(c-v)t/(sqrt(1-v^2/c^2)) ; t'=(1-v/c)t/(sqrt(1-v^2/c^2)) , hence
x'/t'=c.
The front strike light signal moves on the track frame K as x = -ct.
Substituting x in Lorentz equations, we obtain:
x'= - (c+v)t / (sqrt(1-v^2/c^2)) ; t'= (1+ v/c)t / (sqrt(1-v^2/
c^2)) , hence x'/t' = - c.
Then it follows light speed is isotropic in the train frame K’ and has
the same value c as on the track frame K.
Equation on frame K for the back strike light is x_A(t) = ct – 100000
Equation on frame K for the front strike light is x_B(t) = 100000 – ct
So at the track location, on frame K, we have x_A(t) = x_B(t) = 0 ,
which implies t = 100000/c = 1/3 sec. So this is the time when the
track observer sees both strike light signals as simultaneous.
Now, according to the track observer M calculations, the front strike
light signal will reach the train observer M' location (which on the
train frame K’ is x’=0) at point C, while the back strike light signal
will reach the train observer location at point D (both as measured on
frame K), which are given by the following relations:
Point C: train observer M speed relation: x_M = vt = 180000t ; front
strike light signal equation x_B(t) = 100000 - ct. They intersect at
t_C = 100000 / (c + v) = 0.208 sec, at location x_C = vt_C = 37500 km.
On frame K’ of the train observer, point C space-time coordinates are
given by the Lorentz equations, resulting in:
x’_C = (x_C-vt) / (sqrt(1- v^2/c^2)) = (37500 – 180000 * 0.208)/0.8 0 (as it should be)
t’_C = (t_C-vx_C/c^2) / (sqrt(1- v^2/c^2)) = (0.208 – (180000*37500)/
300000^2)/0.8 = 0.167 sec. This is the time, on frame K' when the
front strike light signal reaches the train observer M'.
Point D: train observer M speed relation: x_M = vt = 180000t ; back
strike light signal equation x_B(t) = ct - 100000. They intersect at
t_D = 100000 / (c - v) = 0.833 sec, at location x_D = vt_D = 150000
km.
On frame K’, point D coordinates are given by Lorentz equations,
resulting in:
x’_D = (x_D-vt) / (sqrt(1- v^2/c^2)) = (150000 – 180000 * 0.803)/0.8 0 (as it should)
t’_D = (t_D-vx_D/c^2) / (sqrt(1- v^2/c^2)) = (0.803 – (180000*150000)/
300000^2)/0.8 = 0.667 sec. This is the time, on frame K' when the back
strike light signal reaches the train observer M'.
Conclusions:
a) Track observer on frame K sees two simultaneous strikes at t = 1/3
sec, x = 0.
b) Train observer on frame K’ sees two non-simultaneous strikes, the
first at t’ = 0.167 sec, x’ = 0 and the second at t’ = 0.667 sec at x’
= 0.
This proves the relativity of simultaneity.
Miguel Rios |
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G. L. Bradford
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| Posted: Mon Jul 14, 2008 10:49 pm Post subject: Re: The SR definition led physicists to a 100 years of wild |
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"Dr. Henri Wilson" <HW@....> wrote in message
news:434m74h93vtvgd6m23g38e5ki9hsh2c85p@4ax.com...
| Quote: |
On Fri, 11 Jul 2008 10:49:51 -0700 (PDT), "paparios@gmail.com"
paparios@gmail.com> wrote:
On 11 jul, 10:54, kenseto <kens...@erinet.com> wrote:
On Jul 10, 11:36 am, PD <TheDraperFam...@gmail.com> wrote:
But item #3 is based on the bogus assertion that closing speeds as
perceived by the track observer will affect the isotropy of the
speed
of light in the train.
No, it's based on what he SEES.
Hey idiot he does not see that the strikes were not simultaneous.
Einstein said said that he (the train observer) rush toward the light
from the front and reced away the light from the rear and that's why
he must see the strikes arrive at him at different times. The problem
with this assertion is that it violates the measured isotropy of the
speed of light in the train.
You clearly do not understand English...
Einstein says:
"...When we say that the lightning strokes A and B are simultaneous
with respect to the embankment, we mean: the rays of light emitted at
the places A and B, where the lightning occurs, meet each other at the
mid-point M of the length A -> B of the embankment. But the events A
and B also correspond to positions A and B on the train. Let M' be the
mid-point of the distance A -> B on the travelling train. Just when
the flashes 1 of lightning occur, this point M' naturally coincides
with the point M, but it moves towards the right in the diagram with
the velocity v of the train. If an observer sitting in the position M'
in the train did not possess this velocity, then he would remain
permanently at M, and the light rays emitted by the flashes of
lightning A and B would reach him simultaneously, i.e. they would meet
just where he is situated. Now in reality (considered with reference
to the railway embankment) he is hastening towards the beam of light
coming from B, whilst he is riding on ahead of the beam of light
coming from A. Hence the observer will see the beam of light emitted
from B earlier than he will see that emitted from A...."
Einstein clearly is saying that, with respect to the track frame
(where the strikes were observed as simultaneous at the track observer
position M) the observer M' is moving towards the front strike light
signal and away from the back strike light signal and, naturally that
means M' will see both signals at different times.
Precisely.
Einstein is clearly admitting that the light travels at c+v and c-v in the
train frame.
|
No. Einstein (in A ----> ((A') > (M') < (B')) <-- B) is clearly stating the
principle behind expansion of space and time (between A ----> ((A')...) and
the limits of the constant of the speed of light, c, to overcome said
expansion (A ----> ((A') > (M')...).
GLB
| Quote: |
This not the only indication that Einstein was strongly influenced by BaTh
principles. After all, he conversed with Ritz regularly/.
Miguel Rios
Henri Wilson. ASTC,BSc,DSc(T)
www.users.bigpond.com/hewn/index.htm
All religion involves selling a nonexistant product to gullible
unfortunates. Einstein cleverly exploited this principle with his second
postulate. |
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PD
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| Posted: Tue Jul 15, 2008 3:48 am Post subject: Re: The SR definition led physicists to a 100 years of wild |
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On Jul 14, 6:52 pm, HW@....(Dr. Henri Wilson) wrote:
| Quote: |
On Mon, 14 Jul 2008 04:47:32 -0700 (PDT), PD <TheDraperFam...@gmail.com> wrote:
On Jul 14, 4:24 am, HW@....(Dr. Henri Wilson) wrote:
On Thu, 10 Jul 2008 07:44:23 -0700 (PDT), kenseto <kens...@erinet.com> wrote:
On Jul 9, 7:16 pm, HW@....(Dr. Henri Wilson) wrote:
On Wed, 9 Jul 2008 08:44:47 -0700 (PDT), kenseto <kens...@erinet.com> wrote:
On Jul 8, 12:38 am, HW@....(Dr. Henri Wilson) wrote:
Why not?
The train observer will see the
strikes to be simultaneous or not? SR says that the train observer
will not see the strike to be simultaneous.....based on the closing
velocities as seen by the track observer. Do you agree with that?
The train observer knows his speed wrt the source of the flashes, ie., the
marks on the tracks.
You are an idiot....the sources are at the ends of the train. There is
no relative velocity between the train observer and the ends of the
train.
Don't call me an idiot. I'm the only one here who knows what YOU are talking
about.
You fellows are arguing about something you don't understand at all. You
haven't even defined the light sources.
YOU are now claiming the flashes originate from the train ends....at c wt the
train observer (correct in that case). Draper et al. claim the flashes come
from the track.
No, I do not. I see that you have as much difficulty comprehending
what you read as Seto does.
I was replying to Ken. can't you even read properly now?
|
I was responding to what you said *I* claimed. Can't you even recall
what you said, especially when it's quoted in the previous line?
| Quote: |
What they keep getting wrong is that according to Einstein, the
flashes should move at c in both the track and the train frame.
Yes.
....in which
case they would arrive simultaneously to BOTH observers
No. You CLEARLY do not comprehend what you read.
I was reading Ken's statement.
...which is clearly
impossible.
I have corrected you all by pointing out that TWO sets of reflected flashes
What *reflected* flashes? Reflected from WHAT?
Hey idiot. the marks on the track and train are not the light sources.
Einstein's pathetic experiment uses lightning flashes as sources. They are
presumeably vertical in the track frame and occur instantaneously and
simultaneously in the track frame.
|
Yes.
| Quote: |
The light that reaches the train observer is obviously supposed to be that
which is emitted HORIZONTALLY as the lightning strikes the track.
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Yes.
| Quote: |
That moves at
c wrt the track
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Yes.
| Quote: |
and c+v wrt the train.
|
No.
And you never *did* answer the question, "What *reflected* flashes?
Reflected from WHAT?" You seem to be having difficulty sustaining any
coherency from one post to the next.
| Quote: |
will result from the strikes. One pair move at c wrt the track and the other at
c wrt the train. The train observer receives THREE flashes.
WHAATTT? Now you are babbling. Please, Henri, there is a bit of
spittle hanging off the left corner of your mouth. No, YOUR left.
Draper, you are clearly out of your depth in this NG.
Einstein used lightning flashes because he was an ignorant fool.
What he should have used was a pair of simultaneous horizontal and minute
flashes from sources placed on the track.
|
OK. Why is that materially different than what he did use?
| Quote: |
The light that reaches the train
observer from those sources moves at c in the track frame
|
Yes.
| Quote: |
and c+v and c-v in
the train frame.
|
No.
| Quote: |
Einstein's other claim was to deduce from the marks on both track and train
that two other identical sources were moving at v wrt the track.
|
He made no such statement, nor is there a need to deduce it. Have you
even read what he wrote? If you have not, then how do you know what
the claim was?
| Quote: |
He was correct here.
Lightning struck both track and train simultaneously.
Light was reflected horizontally from both the stationary track and the moving
train ends....ie., there were actually FOUR sources of horizontal light.
|
Ah, first two, then three, now four. Might as well make it seven and
serve cocktails.
| Quote: |
....but we already know that Einstein was a charlaton and that his followers
are all idiots like YOU Draper.
The two flashes of light will not rech the train observer simultaneously, as
Henri Wilson. ASTC,BSc,DSc(T)www.users.bigpond.com/hewn/index.htm
All religion involves selling a nonexistant product to gullible unfortunates. Einstein cleverly exploited this principle with his second postulate. |
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Dr. Henri Wilson
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| Posted: Tue Jul 15, 2008 4:23 am Post subject: Re: The SR definition led physicists to a 100 years of wild |
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On Mon, 14 Jul 2008 12:09:48 -0700 (PDT), "paparios@gmail.com"
<paparios@gmail.com> wrote:
| Quote: |
On 14 jul, 09:27, kenseto <kens...@erinet.com> wrote:
On Jul 10, 11:36 am, PD <TheDraperFam...@gmail.com> wrote:
No, it's based on what he SEES. It is not based on any assumption,
argument, or chain of logic. It is not based on a model, it is not
based on a postulate or principle of relativity. It is the result of
DIRECT OBSERVATION. This is what he SEES with his own EYES.
Hey idiot there is no such observation. Einstein used closing
velocities to conclude that the light from the strikes arrive at the
train observer at different times. Such conclusion is not OBSERVATION
and such conclusion violates the isotropy of the speed of light in the
train. You are so stupid.
Ken Seto
OK, I will put here the complete derivation, proving the relativity of
simultaneity. I'll use some numbers, so you can check the results. I'm
using the same notation as Einstein in http://www.bartleby.com/173/9.html
v= 0.6c = 180000 km/sec
c= 300000 km/sec
sqrt(1-v^2/c^2) = 0.8
Let K be the frame of the track observer M. Also let K’ be the frame
of the train observer M’. The Lorentz transformation equations between
these two frames are:
x'=(x-vt) / (sqrt(1- v^2/c^2)) ; t’ = (t-vx/c^2) / (sqrt(1- v^2/c^2))
When, on frame K, (x,t) = (0,0) it follows that, on frame K’, (x’,t’)
= (0,0) , meaning observers M and M’ coincide. At that instant of time
(t = t' =0), there are two strikes at locations A, at x_A = -100000km,
and location B, at x_B= + 100000km
The back strike light signal moves along the track frame K as x = ct.
Substituting x in Lorentz equations, we obtain:
x'=(c-v)t/(sqrt(1-v^2/c^2)) ; t'=(1-v/c)t/(sqrt(1-v^2/c^2)) , hence
x'/t'=c.
The front strike light signal moves on the track frame K as x = -ct.
Substituting x in Lorentz equations, we obtain:
x'= - (c+v)t / (sqrt(1-v^2/c^2)) ; t'= (1+ v/c)t / (sqrt(1-v^2/
c^2)) , hence x'/t' = - c.
Then it follows light speed is isotropic in the train frame K’ and has
the same value c as on the track frame K.
Equation on frame K for the back strike light is x_A(t) = ct – 100000
Equation on frame K for the front strike light is x_B(t) = 100000 – ct
So at the track location, on frame K, we have x_A(t) = x_B(t) = 0 ,
which implies t = 100000/c = 1/3 sec. So this is the time when the
track observer sees both strike light signals as simultaneous.
Now, according to the track observer M calculations, the front strike
light signal will reach the train observer M' location (which on the
train frame K’ is x’=0) at point C, while the back strike light signal
will reach the train observer location at point D (both as measured on
frame K), which are given by the following relations:
Point C: train observer M speed relation: x_M = vt = 180000t ; front
strike light signal equation x_B(t) = 100000 - ct. They intersect at
t_C = 100000 / (c + v) = 0.208 sec, at location x_C = vt_C = 37500 km.
On frame K’ of the train observer, point C space-time coordinates are
given by the Lorentz equations, resulting in:
x’_C = (x_C-vt) / (sqrt(1- v^2/c^2)) = (37500 – 180000 * 0.208)/0.8 =
0 (as it should be)
t’_C = (t_C-vx_C/c^2) / (sqrt(1- v^2/c^2)) = (0.208 – (180000*37500)/
300000^2)/0.8 = 0.167 sec. This is the time, on frame K' when the
front strike light signal reaches the train observer M'.
Point D: train observer M speed relation: x_M = vt = 180000t ; back
strike light signal equation x_B(t) = ct - 100000. They intersect at
t_D = 100000 / (c - v) = 0.833 sec, at location x_D = vt_D = 150000
km.
On frame K’, point D coordinates are given by Lorentz equations,
resulting in:
x’_D = (x_D-vt) / (sqrt(1- v^2/c^2)) = (150000 – 180000 * 0.803)/0.8 =
0 (as it should)
t’_D = (t_D-vx_D/c^2) / (sqrt(1- v^2/c^2)) = (0.803 – (180000*150000)/
300000^2)/0.8 = 0.667 sec. This is the time, on frame K' when the back
strike light signal reaches the train observer M'.
Conclusions:
a) Track observer on frame K sees two simultaneous strikes at t = 1/3
sec, x = 0.
b) Train observer on frame K’ sees two non-simultaneous strikes, the
first at t’ = 0.167 sec, x’ = 0 and the second at t’ = 0.667 sec at x’
= 0.
This proves the relativity of simultaneity.
|
hahahahahohohohohohohhawhawhahwhahahahaha!
Why go to all that circular trouble?
Just accept light moves at c+v and c-v in the train frame......... and the math
is trivial.
Henri Wilson. ASTC,BSc,DSc(T)
www.users.bigpond.com/hewn/index.htm
All religion involves selling a nonexistant product to gullible unfortunates. Einstein cleverly exploited this principle with his second postulate. |
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Dr. Henri Wilson
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| Posted: Tue Jul 15, 2008 4:36 am Post subject: Re: The SR definition led physicists to a 100 years of wild |
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|
On Mon, 14 Jul 2008 04:41:03 -0700 (PDT), PD <TheDraperFamily@gmail.com> wrote:
| Quote: |
On Jul 14, 3:53 am, HW@....(Dr. Henri Wilson) wrote:
On Fri, 11 Jul 2008 06:10:56 -0700 (PDT), PD <TheDraperFam...@gmail.com> wrote:
On Jul 11, 1:35 am, "G. L. Bradford" <glbra...@insightbb.com> wrote:
"PD" <TheDraperFam...@gmail.com> wrote in message
news:bdd4ad89-b911-4c70-a4b9-d882bc4b6968@2g2000hsn.googlegroups.com...
On Jul 9, 6:12 pm, HW@....(Dr. Henri Wilson) wrote:
us or not simultaneous?
Draper, by the time the light reaches the train obserrver he is NOT at the
midway point.
Of course he is, in the train frame, you idiot. It makes no difference
whatsoever that the track observer is now a mile and half behind by
the time the light reaches the train observer. The train observer
knows he is standing in the middle of the train and verifies at his
leisure that the marks are at the ends of the train. How in this frame
is he not at the midway point? He never moved from the middle of the
train. Idiot.
The train ends are not the source. The lightning is the source.
You are as stupid as Draper.
That WAS Draper, idiot.
The flashes are instantaneous events in the gedanken, idiot.
Instantaneous events are neither at rest nor moving in any frame,
idiot.
|
that's right. The events are instantaneous and simultaneous everywhere.
The sources and their relative speeds are what you should be worrying about
Draper, not your own insanity.
| Quote: |
INTERRUPT
=========================
Relativity: The Special and General Theory, chapter eight, On the Idea of
Time in Physics, by Albert Einstein: "Lightning has struck the rails on our
railway embankment at two places A and B far distant from each other...."
(Lightning has struck the RAILS.....! Two places on the RAILS, not two
places on the train!)
The lightning leaves scorch marks both on the rails and on the train.
Hey idiot, the marks are not the sources. The lightning flshes are.
They are at horizontal rest in the train frame but move at c+v and c-v in the
train frame.
IF the flashes result in secondary emission/reflection from both the track and
the train ends, the train observer will see THREE flashes. One will arrive at
D/(c-v) TWO arrive at D/c and another will arrive at D /(c+v)
GLB
========================
END INTERRUPT
If he performs an elementary calculation, he will establish that the
flashes
occured simultaneously at the two points on the track equidistant from
where he
was at that instant.
The flash sources are assumed at rest in the track frame.
Why on earth would he assume that? He's got no evidence of that. The
flash sources are momentary events at the ENDS OF THE TRAIN. Why on
earth would he assume that they are stationary in the track frame?
Their light moves at
c in that frame and at c+v and c-v in the train observer's frame.
The whole exercise is trivially simple.
Especially to the simple-minded who make stuff up as they go along.
Einstein was a hoaxer.
Just tell me what YOU think, Henri. I'm in the mood for a little
levity.
It's all so simple really.
Ken Seto
Henri Wilson. ASTC,BSc,DSc(T)www.users.bigpond.com/hewn/index.htm
All religion involves selling a nonexistant product to gullible
unfortunates. Einstein cleverly exploited this principle with his
second postulate.
Henri Wilson. ASTC,BSc,DSc(T)www.users.bigpond.com/hewn/index.htm
All religion involves selling a nonexistant product to gullible
unfortunates. Einstein cleverly exploited this principle with his second
postulate.
Henri Wilson. ASTC,BSc,DSc(T)www.users.bigpond.com/hewn/index.htm
All religion involves selling a nonexistant product to gullible unfortunates. Einstein cleverly exploited this principle with his second postulate.
|
Henri Wilson. ASTC,BSc,DSc(T)
www.users.bigpond.com/hewn/index.htm
All religion involves selling a nonexistant product to gullible unfortunates. Einstein cleverly exploited this principle with his second postulate. |
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Dr. Henri Wilson
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| Posted: Tue Jul 15, 2008 4:52 am Post subject: Re: The SR definition led physicists to a 100 years of wild |
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On Mon, 14 Jul 2008 04:47:32 -0700 (PDT), PD <TheDraperFamily@gmail.com> wrote:
| Quote: |
On Jul 14, 4:24 am, HW@....(Dr. Henri Wilson) wrote:
On Thu, 10 Jul 2008 07:44:23 -0700 (PDT), kenseto <kens...@erinet.com> wrote:
On Jul 9, 7:16 pm, HW@....(Dr. Henri Wilson) wrote:
On Wed, 9 Jul 2008 08:44:47 -0700 (PDT), kenseto <kens...@erinet.com> wrote:
On Jul 8, 12:38 am, HW@....(Dr. Henri Wilson) wrote:
Why not?
The train observer will see the
strikes to be simultaneous or not? SR says that the train observer
will not see the strike to be simultaneous.....based on the closing
velocities as seen by the track observer. Do you agree with that?
The train observer knows his speed wrt the source of the flashes, ie., the
marks on the tracks.
You are an idiot....the sources are at the ends of the train. There is
no relative velocity between the train observer and the ends of the
train.
Don't call me an idiot. I'm the only one here who knows what YOU are talking
about.
You fellows are arguing about something you don't understand at all. You
haven't even defined the light sources.
YOU are now claiming the flashes originate from the train ends....at c wt the
train observer (correct in that case). Draper et al. claim the flashes come
from the track.
No, I do not. I see that you have as much difficulty comprehending
what you read as Seto does.
|
I was replying to Ken. can't you even read properly now?
| Quote: |
What they keep getting wrong is that according to Einstein, the
flashes should move at c in both the track and the train frame.
Yes.
....in which
case they would arrive simultaneously to BOTH observers
No. You CLEARLY do not comprehend what you read.
|
I was reading Ken's statement.
| Quote: |
...which is clearly
impossible.
I have corrected you all by pointing out that TWO sets of reflected flashes
What *reflected* flashes? Reflected from WHAT?
|
Hey idiot. the marks on the track and train are not the light sources.
Einstein's pathetic experiment uses lightning flashes as sources. They are
presumeably vertical in the track frame and occur instantaneously and
simultaneously in the track frame.
The light that reaches the train observer is obviously supposed to be that
which is emitted HORIZONTALLY as the lightning strikes the track. That moves at
c wrt the track and c+v wrt the train.
| Quote: |
will result from the strikes. One pair move at c wrt the track and the other at
c wrt the train. The train observer receives THREE flashes.
WHAATTT? Now you are babbling. Please, Henri, there is a bit of
spittle hanging off the left corner of your mouth. No, YOUR left.
|
Draper, you are clearly out of your depth in this NG.
Einstein used lightning flashes because he was an ignorant fool.
What he should have used was a pair of simultaneous horizontal and minute
flashes from sources placed on the track. The light that reaches the train
observer from those sources moves at c in the track frame and c+v and c-v in
the train frame.
Einstein's other claim was to deduce from the marks on both track and train
that two other identical sources were moving at v wrt the track.
He was correct here.
Lightning struck both track and train simultaneously.
Light was reflected horizontally from both the stationary track and the moving
train ends....ie., there were actually FOUR sources of horizontal light.
.....but we already know that Einstein was a charlaton and that his followers
are all idiots like YOU Draper.
| Quote: |
The two flashes of light will not rech the train observer simultaneously, as
|
Henri Wilson. ASTC,BSc,DSc(T)
www.users.bigpond.com/hewn/index.htm
All religion involves selling a nonexistant product to gullible unfortunates. Einstein cleverly exploited this principle with his second postulate. |
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kenseto
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| Posted: Tue Jul 15, 2008 1:51 pm Post subject: Re: The SR definition led physicists to a 100 years of wild |
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On Jul 14, 3:09 pm, "papar...@gmail.com" <papar...@gmail.com> wrote:
| Quote: |
On 14 jul, 09:27, kenseto <kens...@erinet.com> wrote:
On Jul 10, 11:36 am, PD <TheDraperFam...@gmail.com> wrote:
No, it's based on what he SEES. It is not based on any assumption,
argument, or chain of logic. It is not based on a model, it is not
based on a postulate or principle of relativity. It is the result of
DIRECT OBSERVATION. This is what he SEES with his own EYES.
Hey idiot there is no such observation. Einstein used closing
velocities to conclude that the light from the strikes arrive at the
train observer at different times. Such conclusion is not OBSERVATION
and such conclusion violates the isotropy of the speed of light in the
train. You are so stupid.
Ken Seto
OK, I will put here the complete derivation, proving the relativity of
simultaneity. I'll use some numbers, so you can check the results. I'm
using the same notation as Einstein inhttp://www.bartleby.com/173/9.html
v= 0.6c = 180000 km/sec
c= 300000 km/sec
sqrt(1-v^2/c^2) = 0.8
Let K be the frame of the track observer M. Also let K’ be the frame
of the train observer M’. The Lorentz transformation equations between
these two frames are:
x'=(x-vt) / (sqrt(1- v^2/c^2)) ; t’ = (t-vx/c^2) / (sqrt(1- v^2/c^2))
When, on frame K, (x,t) = (0,0) it follows that, on frame K’, (x’,t’)
= (0,0) , meaning observers M and M’ coincide. At that instant of time
(t = t' =0), there are two strikes at locations A, at x_A = -100000km,
and location B, at x_B= + 100000km
The back strike light signal moves along the track frame K as x = ct.
Substituting x in Lorentz equations, we obtain:
x'=(c-v)t/(sqrt(1-v^2/c^2)) ; t'=(1-v/c)t/(sqrt(1-v^2/c^2)) , hence
x'/t'=c.
The front strike light signal moves on the track frame K as x = -ct.
Substituting x in Lorentz equations, we obtain:
x'= - (c+v)t / (sqrt(1-v^2/c^2)) ; t'= (1+ v/c)t / (sqrt(1-v^2/
c^2)) , hence x'/t' = - c.
Then it follows light speed is isotropic in the train frame K’ and has
the same value c as on the track frame K.
Equation on frame K for the back strike light is x_A(t) = ct – 100000
Equation on frame K for the front strike light is x_B(t) = 100000 – ct
So at the track location, on frame K, we have x_A(t) = x_B(t) = 0 ,
which implies t = 100000/c = 1/3 sec. So this is the time when the
track observer sees both strike light signals as simultaneous.
Now, according to the track observer M calculations, the front strike
light signal will reach the train observer M' location (which on the
train frame K’ is x’=0) at point C, while the back strike light signal
will reach the train observer location at point D (both as measured on
frame K), which are given by the following relations:
Point C: train observer M speed relation: x_M = vt = 180000t ; front
strike light signal equation x_B(t) = 100000 - ct. They intersect at
t_C = 100000 / (c + v) = 0.208 sec, at location x_C = vt_C = 37500 km.
On frame K’ of the train observer, point C space-time coordinates are
given by the Lorentz equations, resulting in:
x’_C = (x_C-vt) / (sqrt(1- v^2/c^2)) = (37500 – 180000 * 0.208)/0..8 > 0 (as it should be)
t’_C = (t_C-vx_C/c^2) / (sqrt(1- v^2/c^2)) = (0.208 – (180000*37500)/
300000^2)/0.8 = 0.167 sec. This is the time, on frame K' when the
front strike light signal reaches the train observer M'.
Point D: train observer M speed relation: x_M = vt = 180000t ; back
strike light signal equation x_B(t) = ct - 100000. They intersect at
t_D = 100000 / (c - v) = 0.833 sec, at location x_D = vt_D = 150000
km.
On frame K’, point D coordinates are given by Lorentz equations,
resulting in:
x’_D = (x_D-vt) / (sqrt(1- v^2/c^2)) = (150000 – 180000 * 0.803)/0.8 > 0 (as it should)
t’_D = (t_D-vx_D/c^2) / (sqrt(1- v^2/c^2)) = (0.803 – (180000*150000)/
300000^2)/0.8 = 0.667 sec. This is the time, on frame K' when the back
strike light signal reaches the train observer M'.
Conclusions:
a) Track observer on frame K sees two simultaneous strikes at t = 1/3
sec, x = 0.
b) Train observer on frame K’ sees two non-simultaneous strikes, the
first at t’ = 0.167 sec, x’ = 0 and the second at t’ = 0.667 sec at x’
= 0.
|
NO....The track observer sees the flashes to be simultaneous at
t=100,000/300,000 second=1/3 sec
According to the track observer the train observer sees the flashes to
be simultaneous at:
t'= Gamma*100,000/300,000=1.25*100,000/300000 sec=1.25/3 sec
Your problem is that you use the closing velocities to calculate the
time of arrival of the light fronts from the front and the rear. Such
calculations are wrong. The track observer's calculations must obey
the fact that the speed of light in the train is isotropic and that
closing velocities have no effect on the simultaneity of events in the
train frame.
| Quote: |
This proves the relativity of simultaneity.
|
This proves the relativity of simultaneity is wrong and it violates
the isotropy of the speed of light in the train frame.
Ken Seto |
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kenseto
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| Posted: Tue Jul 15, 2008 2:07 pm Post subject: Re: The SR definition led physicists to a 100 years of wild |
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On Jul 14, 7:23 pm, HW@....(Dr. Henri Wilson) wrote:
| Quote: |
On Mon, 14 Jul 2008 12:09:48 -0700 (PDT), "papar...@gmail.com"
papar...@gmail.com> wrote:
On 14 jul, 09:27, kenseto <kens...@erinet.com> wrote:
On Jul 10, 11:36 am, PD <TheDraperFam...@gmail.com> wrote:
No, it's based on what he SEES. It is not based on any assumption,
argument, or chain of logic. It is not based on a model, it is not
based on a postulate or principle of relativity. It is the result of
DIRECT OBSERVATION. This is what he SEES with his own EYES.
Hey idiot there is no such observation. Einstein used closing
velocities to conclude that the light from the strikes arrive at the
train observer at different times. Such conclusion is not OBSERVATION
and such conclusion violates the isotropy of the speed of light in the
train. You are so stupid.
Ken Seto
OK, I will put here the complete derivation, proving the relativity of
simultaneity. I'll use some numbers, so you can check the results. I'm
using the same notation as Einstein inhttp://www.bartleby.com/173/9.html
v= 0.6c = 180000 km/sec
c= 300000 km/sec
sqrt(1-v^2/c^2) = 0.8
Let K be the frame of the track observer M. Also let K’ be the frame
of the train observer M’. The Lorentz transformation equations between
these two frames are:
x'=(x-vt) / (sqrt(1- v^2/c^2)) ; t’ = (t-vx/c^2) / (sqrt(1- v^2/c^2))
When, on frame K, (x,t) = (0,0) it follows that, on frame K’, (x’,t’)
= (0,0) , meaning observers M and M’ coincide. At that instant of time
(t = t' =0), there are two strikes at locations A, at x_A = -100000km,
and location B, at x_B= + 100000km
The back strike light signal moves along the track frame K as x = ct.
Substituting x in Lorentz equations, we obtain:
x'=(c-v)t/(sqrt(1-v^2/c^2)) ; t'=(1-v/c)t/(sqrt(1-v^2/c^2)) , hence
x'/t'=c.
The front strike light signal moves on the track frame K as x = -ct.
Substituting x in Lorentz equations, we obtain:
x'= - (c+v)t / (sqrt(1-v^2/c^2)) ; t'= (1+ v/c)t / (sqrt(1-v^2/
c^2)) , hence x'/t' = - c.
Then it follows light speed is isotropic in the train frame K’ and has
the same value c as on the track frame K.
Equation on frame K for the back strike light is x_A(t) = ct – 100000
Equation on frame K for the front strike light is x_B(t) = 100000 – ct
So at the track location, on frame K, we have x_A(t) = x_B(t) = 0 ,
which implies t = 100000/c = 1/3 sec. So this is the time when the
track observer sees both strike light signals as simultaneous.
Now, according to the track observer M calculations, the front strike
light signal will reach the train observer M' location (which on the
train frame K’ is x’=0) at point C, while the back strike light signal
will reach the train observer location at point D (both as measured on
frame K), which are given by the following relations:
Point C: train observer M speed relation: x_M = vt = 180000t ; front
strike light signal equation x_B(t) = 100000 - ct. They intersect at
t_C = 100000 / (c + v) = 0.208 sec, at location x_C = vt_C = 37500 km.
On frame K’ of the train observer, point C space-time coordinates are
given by the Lorentz equations, resulting in:
x’_C = (x_C-vt) / (sqrt(1- v^2/c^2)) = (37500 – 180000 * 0.208)/0.8 > >0 (as it should be)
t’_C = (t_C-vx_C/c^2) / (sqrt(1- v^2/c^2)) = (0.208 – (180000*37500)/
300000^2)/0.8 = 0.167 sec. This is the time, on frame K' when the
front strike light signal reaches the train observer M'.
Point D: train observer M speed relation: x_M = vt = 180000t ; back
strike light signal equation x_B(t) = ct - 100000. They intersect at
t_D = 100000 / (c - v) = 0.833 sec, at location x_D = vt_D = 150000
km.
On frame K’, point D coordinates are given by Lorentz equations,
resulting in:
x’_D = (x_D-vt) / (sqrt(1- v^2/c^2)) = (150000 – 180000 * 0.803)/0.8 > >0 (as it should)
t’_D = (t_D-vx_D/c^2) / (sqrt(1- v^2/c^2)) = (0.803 – (180000*150000)/
300000^2)/0.8 = 0.667 sec. This is the time, on frame K' when the back
strike light signal reaches the train observer M'.
Conclusions:
a) Track observer on frame K sees two simultaneous strikes at t = 1/3
sec, x = 0.
b) Train observer on frame K’ sees two non-simultaneous strikes, the
first at t’ = 0.167 sec, x’ = 0 and the second at t’ = 0.667 sec at x’
= 0.
This proves the relativity of simultaneity.
hahahahahohohohohohohhawhawhahwhahahahaha!
Why go to all that circular trouble?
Just accept light moves at c+v and c-v in the train frame......... and the math
is trivial.
|
No light does not move at c+v and c-v in the train frame. Light move
isotropically c_train in the train frame. Light moves isotropically
c_track in the track frame. What this mean is that each observer is
using his clock second to define light speed without taking into
account that clock second in different frame has different duration
contant (absolute time content).
The speed of light in any frame is defined as a constant math ratio as
follows:
Light path length of ruler (299,792,458 m long physically)/the
absolute time content for a clock second co-moving with the ruler.
Ken Seto |
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kenseto
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| Posted: Tue Jul 15, 2008 2:27 pm Post subject: Re: The SR definition led physicists to a 100 years of wild |
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On Jul 14, 9:28 am, PD <TheDraperFam...@gmail.com> wrote:
| Quote: |
On Jul 14, 8:20 am, kenseto <kens...@erinet.com> wrote:
On Jul 10, 11:22 am, PD <TheDraperFam...@gmail.com> wrote:
On Jul 10, 10:07 am, kenseto <kens...@erinet.com> wrote:
On Jul 10, 6:43 am, PD <TheDraperFam...@gmail.com> wrote:
On Jul 9, 4:15 pm, kenseto <kens...@erinet.com> wrote:
On Jul 7, 1:23 pm, PD <TheDraperFam...@gmail.com> wrote:
On Jul 7, 10:26 am, kenseto <kens...@erinet.com> wrote:
On Jul 5, 3:03 pm, PD <TheDraperFam...@gmail.com> wrote:
On Jul 5, 7:45 am, kenseto <kens...@erinet.com> wrote:
On Jul 4, 9:47 am, PD <TheDraperFam...@gmail.com> wrote:
No....he said the train observer M' is rushing toward the light front
from the front and thus he sees the this light front first. He also
said that the train observer is recdeing away from the light front
from the rear and thus he sees this light front at a later time. These
assertions is the baisis of RoS.
No it isn't. It is showing that what the train observer SAW is also
consistent with the laws of physics as known by the track observer.
You do not comprehend what you are reading.
Hey idiot....not according to RoS. RoS says that the train observer is
rushing toward the light front from the front (c+v) and receding from
the light front from the rear (c-v). That means that RoS says that the
speed of light in the train is anisotropic.
That is NOT the statement of the relativity of simultaneity. You don't
comprehend what you are reading.
It is a statement explains how RoS is derived. It is a statement says
that the speed of light in the train is anisotropic.
No, it does NOT say the speed of light is anisotropic in the train.
Yes it does. The train observer is rushing toward the light front from
the front (c+v) and he is receding away from the light from the rear
(c-v). That is anisotropic.
The isotropy of light speed does not have anything to do with the
*closing* speeds that have the values (c+v) and (c-v). There is no law
of physics nor any SR that says the *closing speed* is isotropic >or invariant.
The simultaneity of events is also independent of closing speeds. The
gedanken specified the following conditions:
1. The speed of light in the track and the train frame is isotropic.
2. Both the track observer and the train observer are at equal
distance from the strikes.
3. It was also specified that the track observer sees the strikes to
be simultaneous to begin with.
4. From these conditions the train observer must also see the strikes
to be simultaneous.
You do not know the difference between closing speed and measured
light speed.
It would help if you knew the meaning of the terms you are using.
The *closing* speed of c+v that the track observer sees between the
light from the front and the train observer is NOT the approach speed
of the light measured by the train observer.
So why did you use closing speed to determine whether the train
observer sees the strikes to be simultaneous or not???
I don't. The train observer DOES determine that for himself, using
measured approach speed of light (which is c). That is not the point
of looking at the problem from the track frame. Let me repeat: it is
NOT the purpose of looking at the problem from the track frame to
determine whether the strikes are simultaneous in the train frame.
Einstein did....so are you saying that he wa swrong?
No, he did not. You do not comprehend what you are reading.
The
fact that the strikes are not simultaneous in the train frame is
ALREADY determined from the observations by the train observer.
This is bull shit. The train observer cannot make such determination.
Of course he can. He can SEE the flashes of light, you idiot.
Why? Because he does not know if the strikes were simultaneous or not
simultaneous to begin with.
No, he doesn't have to REASON it out. He OBSERVES things.
The
purpose of looking at the problem from the train frame is to show that
the train observer's account is ALSO consistent with the laws of
physics as known by the track observer. Note that simultaneity is not
a law of physics -- one does not need to have the same answer about
simultaneity for the two frames. The principle of relativity is a
statement about the invariance of the *laws of physics* between
inertial frames, not a statement about the invariance of simultaneity
-- simultaneity is not a law of physics.
You mean RoS is not a law of physics???
You do not even comprehend what I wrote. Please read the above and try
again. Jeez, your command of the English language is pathetic.
I said: "Simultaneity is not a law of physics". You said "You mean ROS
is not a law of physics???"
Do you see what kind of bonehead mistakes you are making?
I agree!!!! It is a buch of
bull shit cook up by Einstein to show that time running at different
rates in different frames. \
Don't you think
that the train observer should do that for himself?
The train observer in
fact measures the approach speed of the light to be c, not c+v.
Right....also he said that the speed of light in his frame is
isotropic and that he is at equal distance from the strikes.. Therefore
he determines that the strikes were simltaneous to begin with.
No, those two conditions are not sufficient for simultaneity. The
ADDITIONAL requirement is that the observer actually observes the
light arriving at the observer at the same time.
Right....Einstein already specified that the track observer sees the
strikes to be simultaneous to begin with. Since the train observer
meets the same conditions as the track observer then the train
observer must also sees the strikes to be simultaneous.
But he DOESN'T because he has his OWN set of observations. He is not
reliant on the track observer's observations, or the track observer's
arguments.
This condition is not
met by the train observer. To recap, there are THREE conditions that
must be met for two events to be simultaneous:
1. The speed of the signal from the events to the observer must be
isotropic.
2. The observer is midway between the two events.
3. The observer actually observes the signal arriving at his location
at the same time.
The first two are met in this case, but not the third.
Both observers meet the third condition.
No sir. READ the gedanken experiment and try to comprehend what you
are reading. It says in black and white that the train observer does
NOT meet the third condition. He in fact observes the OPPOSITE.
The train observer does not observe whether the strikes are
simultaneous or not....
Yes, he does. READ the gedanken.
he does not have enough data to make such
observation.
Yes, he does. You've already (just yesterday!) asked me what
observations he made, and I told you, and you've already (just
overnight) forgotten what I told you. Your mind is gone, Seto. You
cannot remember a discussion from one day to the next. I'm about to
give up on you, as your senility is too deep a handicap to overcome.
What he observes are as follows:
1. that the strikes happened at equal distance from him.
2. That the speed of light in his frame is isotropic.
And he observes one more thing which you keep forgetting, day after
day after day, even though it is printed in black and white in the
book you refuse to look up.
3. That the light from the two flashes arrives at his eyes at
different times.
Because he has this third observation, he needs nothing else to
determine the simultaneity of the strikes. In particular, he does NOT
need the chain of logic from the track observer to make any conclusion
about it.
Hey idiot there is no such observation. The light fronts do not arrive
at different times. If they did the speed of light in the train is not
isotropic.
So you can't even follow what is said in a three-minute animation,
where it is plainly said, both in words and in pictures, that the
train observer does in fact make this observation, and that the speed
of the light in the train is in fact isotropic?
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You are a lying sack of shit. Einstein did not say that the train
observer makes this observation. He said that the train observer is
rushing toward the light front from the front and receding away from
the light font from the rear and thus he concluded that the train
observer sees the light front from the front before he sees the light
front from the rear. The problems with his arguement are as follows:
1. What he asserted violates the isotropy of the speed of light in the
train.
2. What he asserted is based on closing velocities between the train
observer and the light fronts. Closing velocities are perceived by the
track observer they have no effect on the simultaneity of events or
the isotropy of the speed of light in the train.
3. The train observer must make his own determination whether the
strikes are simultaneous. Since the track observer determined that the
strikes were simultaneous to begin with then the train observer must
also see the strikes to be simultaneous because he was at equal
distance from the strikes when they occur simultaneously.
Ken Seto
| Quote: |
Since it's a free web resource, you can watch and listen to it as many
times as you like, Ken. AND it's based EXACTLY on what Einstein laid
out in his gedanken.
I can no longer help you, Ken. Every morning you revert to where you
were yesterday morning, having lost all that you learned yesterday
afternoon.
PD- Hide quoted text -
- Show quoted text - |
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paparios@gmail.com
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| Posted: Tue Jul 15, 2008 2:35 pm Post subject: Re: The SR definition led physicists to a 100 years of wild |
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On 15 jul, 09:51, kenseto <kens...@erinet.com> wrote:
| Quote: |
On Jul 14, 3:09 pm, "papar...@gmail.com" <papar...@gmail.com> wrote:
On 14 jul, 09:27, kenseto <kens...@erinet.com> wrote:
On Jul 10, 11:36 am, PD <TheDraperFam...@gmail.com> wrote:
No, it's based on what he SEES. It is not based on any assumption,
argument, or chain of logic. It is not based on a model, it is not
based on a postulate or principle of relativity. It is the result of
DIRECT OBSERVATION. This is what he SEES with his own EYES.
Hey idiot there is no such observation. Einstein used closing
velocities to conclude that the light from the strikes arrive at the
train observer at different times. Such conclusion is not OBSERVATION
and such conclusion violates the isotropy of the speed of light in the
train. You are so stupid.
Ken Seto
OK, I will put here the complete derivation, proving the relativity of
simultaneity. I'll use some numbers, so you can check the results. I'm
using the same notation as Einstein inhttp://www.bartleby.com/173/9.html
v= 0.6c = 180000 km/sec
c= 300000 km/sec
sqrt(1-v^2/c^2) = 0.8
Let K be the frame of the track observer M. Also let K’ be the frame
of the train observer M’. The Lorentz transformation equations between
these two frames are:
x'=(x-vt) / (sqrt(1- v^2/c^2)) ; t’ = (t-vx/c^2) / (sqrt(1- v^2/c^2))
When, on frame K, (x,t) = (0,0) it follows that, on frame K’, (x’,t’)
= (0,0) , meaning observers M and M’ coincide. At that instant of time
(t = t' =0), there are two strikes at locations A, at x_A = -100000km,
and location B, at x_B= + 100000km
The back strike light signal moves along the track frame K as x = ct.
Substituting x in Lorentz equations, we obtain:
x'=(c-v)t/(sqrt(1-v^2/c^2)) ; t'=(1-v/c)t/(sqrt(1-v^2/c^2)) , hence
x'/t'=c.
The front strike light signal moves on the track frame K as x = -ct.
Substituting x in Lorentz equations, we obtain:
x'= - (c+v)t / (sqrt(1-v^2/c^2)) ; t'= (1+ v/c)t / (sqrt(1-v^2/
c^2)) , hence x'/t' = - c.
Then it follows light speed is isotropic in the train frame K’ and has
the same value c as on the track frame K.
Equation on frame K for the back strike light is x_A(t) = ct – 100000
Equation on frame K for the front strike light is x_B(t) = 100000 – ct
So at the track location, on frame K, we have x_A(t) = x_B(t) = 0 ,
which implies t = 100000/c = 1/3 sec. So this is the time when the
track observer sees both strike light signals as simultaneous.
Now, according to the track observer M calculations, the front strike
light signal will reach the train observer M' location (which on the
train frame K’ is x’=0) at point C, while the back strike light signal
will reach the train observer location at point D (both as measured on
frame K), which are given by the following relations:
Point C: train observer M speed relation: x_M = vt = 180000t ; front
strike light signal equation x_B(t) = 100000 - ct. They intersect at
t_C = 100000 / (c + v) = 0.208 sec, at location x_C = vt_C = 37500 km.
On frame K’ of the train observer, point C space-time coordinates are
given by the Lorentz equations, resulting in:
x’_C = (x_C-vt) / (sqrt(1- v^2/c^2)) = (37500 – 180000 * 0.208)/0.8 > > 0 (as it should be)
t’_C = (t_C-vx_C/c^2) / (sqrt(1- v^2/c^2)) = (0.208 – (180000*37500)/
300000^2)/0.8 = 0.167 sec. This is the time, on frame K' when the
front strike light signal reaches the train observer M'.
Point D: train observer M speed relation: x_M = vt = 180000t ; back
strike light signal equation x_B(t) = ct - 100000. They intersect at
t_D = 100000 / (c - v) = 0.833 sec, at location x_D = vt_D = 150000
km.
On frame K’, point D coordinates are given by Lorentz equations,
resulting in:
x’_D = (x_D-vt) / (sqrt(1- v^2/c^2)) = (150000 – 180000 * 0.803)/0.8 > > 0 (as it should)
t’_D = (t_D-vx_D/c^2) / (sqrt(1- v^2/c^2)) = (0.803 – (180000*150000)/
300000^2)/0.8 = 0.667 sec. This is the time, on frame K' when the back
strike light signal reaches the train observer M'.
Conclusions:
a) Track observer on frame K sees two simultaneous strikes at t = 1/3
sec, x = 0.
b) Train observer on frame K’ sees two non-simultaneous strikes, the
first at t’ = 0.167 sec, x’ = 0 and the second at t’ = 0.667 sec at x’
= 0.
NO....The track observer sees the flashes to be simultaneous at
t=100,000/300,000 second=1/3 sec
According to the track observer the train observer sees the flashes to
be simultaneous at:
t'= Gamma*100,000/300,000=1.25*100,000/300000 sec=1.25/3 sec
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So you are not using SR there. Your math is wrong. Mine is right!
| Quote: |
Your problem is that you use the closing velocities to calculate the
time of arrival of the light fronts from the front and the rear. Such
calculations are wrong. The track observer's calculations must obey
the fact that the speed of light in the train is isotropic and that
closing velocities have no effect on the simultaneity of events in the
train frame.
|
Those calculation are not wrong since they are done IN THE TRACK
FRAME, which is the frame who it is being used as reference. For the
track observer the closing speed DO EXIST.
| Quote: |
This proves the relativity of simultaneity.
This proves the relativity of simultaneity is wrong and it violates
the isotropy of the speed of light in the train frame.
Ken Seto
|
It does not. I carefully proved the speed of light to be isotropic in
the train frame, by using Lorentz equations. Same equations were used
to determine arrival times to the train observer.
Miguel Rios |
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PD
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| Posted: Tue Jul 15, 2008 3:25 pm Post subject: Re: The SR definition led physicists to a 100 years of wild |
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On Jul 15, 9:27 am, kenseto <kens...@erinet.com> wrote:
| Quote: |
On Jul 14, 9:28 am, PD <TheDraperFam...@gmail.com> wrote:
On Jul 14, 8:20 am, kenseto <kens...@erinet.com> wrote:
On Jul 10, 11:22 am, PD <TheDraperFam...@gmail.com> wrote:
On Jul 10, 10:07 am, kenseto <kens...@erinet.com> wrote:
On Jul 10, 6:43 am, PD <TheDraperFam...@gmail.com> wrote:
On Jul 9, 4:15 pm, kenseto <kens...@erinet.com> wrote:
On Jul 7, 1:23 pm, PD <TheDraperFam...@gmail.com> wrote:
On Jul 7, 10:26 am, kenseto <kens...@erinet.com> wrote:
On Jul 5, 3:03 pm, PD <TheDraperFam...@gmail.com> wrote:
On Jul 5, 7:45 am, kenseto <kens...@erinet.com> wrote:
On Jul 4, 9:47 am, PD <TheDraperFam...@gmail.com> wrote:
No....he said the train observer M' is rushing toward the light front
from the front and thus he sees the this light front first. He also
said that the train observer is recdeing away from the light front
from the rear and thus he sees this light front at a later time. These
assertions is the baisis of RoS.
No it isn't. It is showing that what the train observer SAW is also
consistent with the laws of physics as known by the track observer.
You do not comprehend what you are reading.
Hey idiot....not according to RoS. RoS says that the train observer is
rushing toward the light front from the front (c+v) and receding from
the light front from the rear (c-v). That means that RoS says that the
speed of light in the train is anisotropic.
That is NOT the statement of the relativity of simultaneity. You don't
comprehend what you are reading.
It is a statement explains how RoS is derived. It is a statement says
that the speed of light in the train is anisotropic.
No, it does NOT say the speed of light is anisotropic in the train.
Yes it does. The train observer is rushing toward the light front from
the front (c+v) and he is receding away from the light from the rear
(c-v). That is anisotropic.
The isotropy of light speed does not have anything to do with the
*closing* speeds that have the values (c+v) and (c-v). There is no law
of physics nor any SR that says the *closing speed* is isotropic >or invariant.
The simultaneity of events is also independent of closing speeds. The
gedanken specified the following conditions:
1. The speed of light in the track and the train frame is isotropic.
2. Both the track observer and the train observer are at equal
distance from the strikes.
3. It was also specified that the track observer sees the strikes to
be simultaneous to begin with.
4. From these conditions the train observer must also see the strikes
to be simultaneous.
You do not know the difference between closing speed and measured
light speed.
It would help if you knew the meaning of the terms you are using.
The *closing* speed of c+v that the track observer sees between the
light from the front and the train observer is NOT the approach speed
of the light measured by the train observer.
So why did you use closing speed to determine whether the train
observer sees the strikes to be simultaneous or not???
I don't. The train observer DOES determine that for himself, using
measured approach speed of light (which is c). That is not the point
of looking at the problem from the track frame. Let me repeat: it is
NOT the purpose of looking at the problem from the track frame to
determine whether the strikes are simultaneous in the train frame.
Einstein did....so are you saying that he wa swrong?
No, he did not. You do not comprehend what you are reading.
The
fact that the strikes are not simultaneous in the train frame is
ALREADY determined from the observations by the train observer.
This is bull shit. The train observer cannot make such determination.
Of course he can. He can SEE the flashes of light, you idiot.
Why? Because he does not know if the strikes were simultaneous or not
simultaneous to begin with.
No, he doesn't have t | | | |
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